x.4(2x+1)=3(x-1)+7

Solution for x.4(2x+1)=3(x-1)+7 equation:

x.4(2x+1)=3(x-1)+7

We move all terms to the left:

x.4(2x+1)-(3(x-1)+7)=0

We multiply parentheses

2x^2+x-(3(x-1)+7)=0


We calculate terms in parentheses: -(3(x-1)+7), so:

3(x-1)+7

We multiply parentheses

3x-3+7

We add all the numbers together, and all the variables

3x+4

Back to the equation:

-(3x+4)


We get rid of parentheses

2x^2+x-3x-4=0

We add all the numbers together, and all the variables

2x^2-2x-4=0

a = 2; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·2·(-4)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*2}=\frac{-4}{4}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*2}=\frac{8}{4}
=2
$